\chapter{Transfer Impedance model according to Kley} \label{appendix_kley}

From the work of Vance \cite{Vance} and Kley \cite{Kley} and the summary of this work in \cite{Tesche} we may calculate the frequency dependent transfer impedance of a braided wire shield from the cable shield parameters. The shield parameters are 

\begin{enumerate}
\item braid diameter, D 
\item Number of carriers, C
\item Number of wires in a carrier, N
\item diameter of a single wire, d 
\item conductivity of wires, $\sigma$
\item pitch angle of the braid, $\alpha$
\end{enumerate}

These parameters are illustrated in figure \ref{fig:braid2}.

\begin{figure}[h]
\centering
\includegraphics[scale=0.25]{./Imgs/braid.jpg}
\caption{Braid parameters}
\label{fig:braid2}    
\end{figure}

The transfer impedance is the sum of the effects of diffusion through the shield conductor, penetration of the field through the holes in the shield and the effects of the overalpping weave of the braid conductors. 
\begin{equation} \label{eq:Kley_0}
Z_t=Z_d+j\omega M_h +j\omega M_b
\end{equation}

The calculation of the transfer impedance terms proceeds as follows:

The fill factor, F, is calculated as
%
\begin{equation} \label{eq:Kley_1}
F=\frac{N C d}{2 \pi D cos(\alpha)}
\end{equation}
%
The optical coverage, K, is
%
\begin{equation} \label{eq:Kley_2}
K=2F-F^2
\end{equation}
%
The length and width of the rhombic holes are given by
\begin{equation} \label{eq:Kley_3}
l=\frac{\left(1-F\right)Nd}{Fsin(\alpha)}
\end{equation}
\begin{equation} \label{eq:Kley_4}
w=\frac{\left(1-F\right)Nd}{Fcos(\alpha)}
\end{equation}
%
a parameter, e, is defined as
\begin{equation} \label{eq:Kley_5}
e=\sqrt{1-\left(\frac{w}{l} \right)^2}
\end{equation}

The diffusion impedance term is given by
\begin{equation} \label{eq:Kley_6}
Z_d=R_0\frac{\gamma}{sinh(\gamma t)}
\end{equation}
where $R_0$ is the d.c. resistance of the shield
\begin{equation} \label{eq:Kley_7}
R_0=\frac{4}{\pi d^2 N C \sigma cos(\alpha) }
\end{equation}
$\gamma$ is the propgation constant in the shield conductor
\begin{equation} \label{eq:Kley_8}
\gamma=\frac{1+j}{\delta }
\end{equation}
where
\begin{equation} \label{eq:Kley_9}
\delta=\sqrt{ \frac{2}{\omega \mu_0 \sigma } }
\end{equation}

and $t$ is an equivalent thickness of the cable shield given by
\begin{equation} \label{eq:Kley_10}
t=\frac{1}{\pi \sigma D R_0 }
\end{equation}
    
The hole inductance term is given by   
\begin{equation} \label{eq:Kley_11}
   M_h=\frac{\mu_0 \pi (1-K)^{\frac{3}{2}})e^2}{6 C (Em(e)-(1-e^2)Km(e))}*C_k    
\end{equation}
where $Km$ and $Em$ are elliptic integrals and 
\begin{equation} \label{eq:Kley_12}
C_k=0.875 e^{-T_h} 
\end{equation}
\begin{equation} \label{eq:Kley_13}
T_h=9.6 F \left( \frac{K^2 d}{2a} \right) ^{\frac{1}{3}}
\end{equation}
\begin{equation} \label{eq:Kley_14}
a=D/2d0+d
\end{equation}

    
\begin{equation} \label{eq:Kley_15}
M_b=\frac{-0.22 \mu_0 d}{ 4 \pi D_m F cos( \alpha )}cos(2 k_1 \alpha) 
\end{equation}
where 
\begin{equation} \label{eq:Kley_16}
k_1=\frac{\pi}{2.667 F cos(\alpha)+\frac{\pi}{10} }
\end{equation}
and
\begin{equation} \label{eq:Kley_17}
D_m=D+2d
\end{equation}


\clearpage